(Note that the parts $\lambda_k$ are the arguments on the main diagonal; the partition is padded with zeros to get length $n$.)

Then an interesting identity is implied in this question: if $g(z):=\dfrac1{\Gamma(z+1)}=\dfrac1{z!}$, then we have $$\boxed{\sum\limits_{\lambda\vdash n}\det[M_{fg}(\lambda)\cdot M_{g}(\lambda)] =\dfrac {f(1)^n}{n!}=f(1)^ng(n)},$$ meaning that all other $f(k)$ for $k\ne1$ are canceled out in the summing.
The order of the $\lambda_k$'s is crucial here. Even if all partitions are ordered "backwards", i.e. increasingly, nothing cancels out.